Let $(X,d)$ be a metric space. Assume that $(x_n)$ is a sequence in $X$. There is a well-known theorem about the convergence of $(x_n)$ that reads as follows.
Theorem. Every sub-sequence of $(x_n)$ has a sub-sub-sequence of $(x_n)$ that convergence to $x$ if and only if the sequence $(x_n)$ converges to $x$.
The proof for $\impliedby$ is easy and immediate. For $\implies$, there are proofs in this post which are using contradiction. I was wondering if there is a direct proof for $\implies$?
It is not too difficult to adapt the proof in the case of the real numbers that was linked to in the comments to work in an arbitrary metric space.
The key is to recall that $x_n \to x$ in $X$ if and only if $d(x_n,x) \to 0$ in $\mathbb{R}$.
Now it is clear that $\liminf_{n \to \infty} d(x_n,x) = 0$ since $x_n$ certainly has a subsequence converging to $x$. Additionally, there is a subsequence $x_{n_k}$ such that $d(x_{n_k}, x) \to \limsup_{n \to \infty} d(x_n,x)$ as $k \to \infty$. By passing to a further subsequence, we can assume that $x_{n_k} \to x$ in $X$. This means that $d(x_{n_k},x) \to 0$ so that $\limsup_{n \to \infty} d(x_n,x) = 0$.
Therefore $\lim_{n \to \infty} d(x_n,x) = 0$ as desired.
Here's an elementary argument that holds whenever $ X $ is a topological space.
For any subset $ J \subset \mathbb N $, we define $ X_J = \{ x_n : n \in J \} $. Given an infinite subset $ J \subset \mathbb N $, we can identify a subsequence $ ( x_{ n_k } ) $ of $ ( x_n ) $ with each $ n_k $ coming from $ J $. In particular, the subsequence $ ( x_{ n_k } ) $ lies in $ X_J $. Applying the hypothesis, there exists a further subsequence $ ( x_{ n_{ k_l }} ) $ converging to $ x $, from which we conclude that $ x $ lies in $ \overline{ X_J} $, the closure of $ X_J $. This establishes that $$ J \text{ infinite} \quad \Longrightarrow \quad x \in \overline{ X_J} . $$
Fix now a neighborhood $ U $ of $ x $ and consider the set $ J = \{ n \in \mathbb N : x_n \notin U \} $. Since $ X_J = \{ x_n : x_n \notin U \} $, the elements of $ X_J $ lie outside of $ U $, so they cannot get close to $ x $. More precisely, $ x $ does not lie in $ \overline{ X_J } $, and by the above, the set $ J $ must be finite. Consequently, every neighborhood $ U $ of $ x $ contains all but finitely many terms of the sequence, or the sequence $ ( x_n ) $ converges to $ x $.
If by a "direct proof" you mean a constructive proof, then no; there is no constructive proof.
Precisely, given a metric space $S$, define
Sequence Lemma for $S$: Let $\{c_n\}_{n \in \mathbb{N}}$ be a sequence in $S$ and $L \in S$ such that every subsequence of $c$ has a further subsequence converging to $L$. Then $c$ converges, necessarily to $L$.
In particular, the following statement is not constructively provable:
Limited Principle of Omniscience (LPO): Let $\{b_n\}_{n \in \mathbb{N}}$ be a sequence such that $b_n \in \{0, 1\}$ for all $n$. There either exists $n$ such that $b_n = 1$, or there does not exist such $n$.
I claim that the Sequence Lemma for any metric space with at least $2$ distinct points implies LPO. WLOG, let the two distinct points in $S$ be $0, 1$, distance 1 apart. Consider a sequence $\{b_n\}_{n \in \mathbb{N}}$ taking values in $\{0, 1\}$. Define
$$c_n = \begin{cases} 1 & c_n = 1 \text{ and for all } m < n, c_m = 0 \\\\ 0 & otherwise \end{cases}$$
Note there exists $n$ such that $c_n = 1$ if and only if there exists $n$ such that $b_n = 1$. For if $c_n = 1$, then $b_n = 1$. And if $b_n = 1$, let $k$ be the least $k$ such that $b_k = 1$; then $c_k = 1$.
I claim that every subsequence $d$ of $c$ has a further subsequence $f$ which converges to $0$. Indeed, note there is at most one $n$ such that $c_n = 1$, with all other values $0$. Thus, if $d$ is a subsequence of $c$, there is at most one $n$ such that $d_n = 1$, with all other values $0$. Define
$$n_k = \begin{cases} k & \text{for all } m \leq k, d_m = 0 \\\\ k + 1 & otherwise \end{cases}$$
Then let $f_k = d_{n_k}$. It is straightforward to see that all values of $f$ are $0$, so $\lim\limits_{k \to \infty} f_k = 0$.
Now we have $\lim\limits_{n \to \infty} c_n = 0$. Take $N$ such that for all $n \geq N$, $|c_n| < 1$. Then for all $n \geq N$, $c_n = 0$. Then either there is $k < N$ such that $c_k = 1$, or there is no $k < N$ such that $c_k = 1$. In the latter case, there is no $k$ such that $c_k$, since we have ruled out the cases $k \geq N$ and $k < N$. So there either is $k$ such that $c_k = 1$, or there is no such $k$. This completes the proof that your sequence theorem implies LPO. $\square$
In fact, your sequence theorem is equivalent to LPO under extremely mild choice assumptions. In particular, LPO constructively implies the Sequence Lemma for all metric spaces $S$, assuming a very weak version of countable choice.
For assume LPO, fix $S$, and take $\epsilon > 0$. Suppose we have a sequence $c$ such that every subsequence of $c$ has a subsequence converging to $x$, and fix $\epsilon > 0$. Using an extremely weak version of countable choice, define a sequence $b : \mathbb{N} \to \{0, 1\}$ such that if $b_n = 0$, then $d(c_n, L) < \epsilon$, and if $b_n = 1$, then $d(c_n, L) > \epsilon / 2$. Define
$$d_n = \begin{cases} 0 & \text{there exists } m > n \text{ such that } b_n = 1 \\\\ 1 & otherwise\end{cases}$$
$d$ is well-defined because deciding whether there exists $m > n$ such that $b_m = 1$ is equivalent to deciding whether there exists $k$ such that $b_{n + k + 1} = 1$, which is possible by LPO.
By LPO, there either exists $N$ such that $d_N = 1$, or there does not. If there exists $N$ such that $d_N = 1$, then for all $m > N$, $b_n = 0$ and thus $d(c_n, L) < \epsilon$, as desired. If no such $N$ exists, then recursively define $f_n$ to be the smallest $M$ such that $M > f_0, \ldots, f_{n - 1}$ and $b_n = 1$. Then $f$ is a strictly increasing sequence of natural numbers. Consider the subsequence $w_k = c_{f_k}$; then for all $k$, $d(c_n, L) > \epsilon / 2$, so $w$ has no subsequence converging to $L$; contradiction. This completes the proof of LPO. $\square$
